6 deleted 2 characters in body

I hope this problem is not considered too "elementary" for MO. It concerns a formula that I have always found fascinating. For, at first glance, it appears completely "obvious", while on closer examination it does not even seem well-defined. The formula is the one that I was given as the definition of the cross-product in $\mathbb R^3$ when I was first introduced to that concept:

$$B \times C := \det \begin{vmatrix} {\mathbf i } & {\mathbf j } & {\mathbf k } \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3\\ \end{vmatrix}$$ On the one hand, if one expands this by minors of the first row, the result is clearly correct---and to this day this is the only way I can recall the formula for the components of the cross-product when I need it. But, on the other hand, the determinant of an $n \times n$ matrix whose elements are a mixture of scalars and vectors is undefined. Just think what happens if you interchange one element of the first row with the element just below it. In fact, as usually understood, for a determinant of a matrix to be well-defined, its elements should all belong to a commutative ring. But then again (on the third hand :-) if we take the dot product of both sides of the formula with a third vector( A), $A$, we seem to get:

$$A \cdot B \times C = A \cdot \det \begin{vmatrix} {\mathbf i } & {\mathbf j } & {\mathbf k } \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3\\ \end{vmatrix} = \det \begin{vmatrix} A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3\\ \end{vmatrix}$$ and of course the left and right hand sides are well-known formulas for the (signed) volume of the parallelepiped spanned by the three vectors, $A, B, C$. Moreover, the validity of the latter formula for all choices of $A$ indicates that the original formula is "correct".

So, my question is this: Is there a rigorous way of defining the original determinant so that all of the above becomes meaningful and correct?

5 deleted 2 characters in body

I hope this problem is not considered too "elementary" for MO. It concerns a formula that I have always found fascinating. For, at first glance, it appears completely "obvious", while on closer examination it does not even seem well-defined. The formula is the one that I was given as the definition of the cross-product in $\mathbb R^3$ when I was first introduced to that concept:

$$B \times C := \det \begin{vmatrix} {\mathbf i } & {\mathbf j } & {\mathbf k } \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3\\ \end{vmatrix}$$ On the one hand, if one expands this by minors of the first row, the result is clearly correct---and to this day this is the only way I can recall the formula for the components of the cross-product when I need it. But, on the other hand, the determinant of an ($n \times n) n$ matrix whose elements are a mixture of scalars and vectors is undefined. Just think what happens if you interchange one element of the first row with the element just below it. In fact, as usually understood, for a determinant of a matrix to be well-defined, its elements should all belong to a commutative ring. But then again (on the third hand :-) if we take the dot product of both sides of the formula with a third vector ( A), we seem to get:

$$A \cdot B \times C = A \cdot \det \begin{vmatrix} {\mathbf i } & {\mathbf j } & {\mathbf k } \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3\\ \end{vmatrix} = \det \begin{vmatrix} A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3\\ \end{vmatrix}$$ and of course the left and right hand sides are well-known formulas for the (signed) volume of the parallelepiped spanned by the three vectors, $A, B, C$. Moreover, the validity of the latter formula for all choices of $A$ indicates that the original formula is "correct".

So, my question is this: Is there a rigorous way of defining the original determinant so that all of the above becomes meaningful and correct?